Int. Journal of Math. Analysis, Vol. 1, 2007, no. 24, 1189 - 1208
Positive Solutions for Nonlinear Singular
Boundary Value Problems on the Half Line
Baoqiang Yan1
Department of Mathematics
Shandong Normal University
Ji-nan, 250014, P. R. China
yanbqcn@yahoo.com.cn
Richard M. Timoney2
School of Mathematics
Trinity College
Dublin 2, Ireland
richardt@maths.tcd.ie
Abstract
We discuss the existence of positive solutions for singular secondorder boundary value problems x = µf (t, x, x ), ax(0) − bx (0) = k ≥
0, x (∞) = 0, where f may be singular at x = 0 and x = 0 and
can change sign. Via fixed point theory, we establish the existence of
positive solutions under some conditions on f . Our results deal with the
situation where the solutions approach the singularities of the equation.
Mathematics Subject Classification: 34B16, 34B18, 34B40
Keywords: Fixed point theory; sign-changing equation, positive solution
1
Project supported by the fund of National Natural Science (10571111) and the fund
of Natural Science of Shandong Province (Y2005A07) and carried out at Trinity College
Dublin.
2
Based in part upon work supported by the Science Foundation Ireland under grant
05/RFP/MAT0033.
1190
1
B. Yan and R. M. Timoney
Introduction
Our aim is to establish existence of (nonnegative increasing) solutions to the
boundary value problem
t ∈ (0, +∞)
x (t) = μf (t, x, x ),
lim x (t) = 0
ax(0) − b lim x (t) = k,
t→0
t→∞
(1.1)
(1.2)
where a > 0, b ≥ 0, k ≥ 0, μ > 0, f (t, x, z) is singular at x = 0 and z = 0 and
may change sign. Our result establishes this existence for sufficiently small
values of μ and with relatively few conditions on f (a certain bound on |f | and
a negativity condition on f (t, x, z) for small z). A parameter such as μ can be
called a Thiele modulus, though this terminology is usually applied when the
interval is finite [4, 5].
Our basic method is to apply fixed point results for cones in Banach spaces
to get existence, but we have to apply this method to a sequence of nonsingular
integral equation problems that approximate the problem of interest. The
choice of the approximate problems is somewhat delicate as is the process of
establishing that the solutions to the approximate problems converge.
There is considerable literature which is relevant to our work. Staněk [14]
considers x = μq(t)f (t, x, x ) (0 < t < T ) with boundary conditions of the
type ax(0) − bx (0) = k > 0, x(T ) = 0. The methods of [14] are quite similar
in outline to ours, despite the finiteness of the interval, but [14] requires the
functions q and f to be nonnegative (along with several other conditions).
Via the technique of upper and lower solutions Agarwal and O’Regan [2, 3]
discussed the existence of bounded solutions to equations
1
(p(t)x ) = f (t, x, p(t)x )
p(t)
(0 < t < ∞)
(1.3)
subject to a boundary condition ax(0) − b limt→0 x (t) = c0 . In the case p(t) ≡
1, A. Constantin [6] established the existence of bounded solutions to equations
of the same kind (subject to certain growth conditions on f ). In [2, 3, 6] the
function f (t, x, z) is always assumed continuous for (x, z) ∈ R2 . In [15], again
with p(t) ≡ 1, G. Yang considered positive solutions to (1.3) with boundary
conditions x(0) = x (∞) = 0, allowing f (t, x, z) to be singular at x = 0 and
z = 0 but requiring boundedness of f as x → +∞. See also the general
references [12, 13] and the works [8, 10, 11, 16] but note that [10, 11, 16] do
not consider singular equations.
Theorem 1.1. Assume f ∈ C([0, +∞) × (0, +∞) × (0, +∞), R) and there
exist positive Φ ∈ C[0, +∞),
1 h ∈ C(0, +∞) and g ∈ C(0, +∞) with Φ∞ =
supt∈[0,+∞) |Φ(t)| < +∞, 0 h(s) ds < +∞ and
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Positive solutions
∞
0
Φ(s) max 1 ≤x≤c(1+s) h(x) ds < +∞ for each c ≥ 1. Suppose
c
|f (t, y, z)| ≤ Φ(t)h(y)g(z)
(t > 0, y > 0, z > 0).
(1.4)
Assume also there is β ∈ C(0, +∞), β(t) < 0 and constants 0 ≤ γ < 1, δ > 0
so that
f (t, x, z) ≤ xγ β(t)
(t ≥ 0, x > 0, 0 < z ≤ δ).
(1.5)
Then there is μ0 > 0 so that for each μ in the range 0 < μ ≤ μ0 (1.1)–(1.2)
has a solution x = x0 with x0 (t) ≥ 0 and x0 (t) > 0 (for all t ≥ 0).
Our result implies the result of [15] but in contrast to [15], f may be
superlinear at x = +∞. In contrast to [3] we allow f to have singularities and
our hypotheses admit unbounded solutions. Our result does not generalise
several results from the literature where singularities are not considered such
as [6, Theorem 1]. In the proof of [6, Theorem 4], a specific singular equation
on [0, ∞) is considered where f (t, x, z) is monotone in z and this special feature
of the equation is used in establishing the existence of solutions. In addition
the solution obtained is bounded below by a positive constant and so does not
approach the singularity at x = 0.
Some of our methods are inspired by [1] and [15]. As corollaries, we discuss boundedness and unboundedness of positive solutions and we give some
concrete examples where our methods allow us to give a specific value for μ0 .
2
Preliminaries
We will consider two spaces of functions in addition to the space C(0, +∞) of
continuous (R-valued) functions on the open interval (0, +∞) (and similarly
for other intervals in R). One is
C[0, ∞] = x ∈ C[0, ∞) : x(∞) = lim x(t) exists ,
t→+∞
(the continuous functions on the one point compactification of [0, ∞)) normed
1
{x ∈
by the usual supremum x∞ = supt≥0 |x(t)|. The
other is C∞ [0, ∞) = C[0, ∞) : x ∈ C[0, ∞]} normed by x = max supt≥0 |x(t)|
, supt≥0 |x (t)| . It
1+t
is easy to verify that the existence of x (∞) = limt→∞ x (t) implies finiteness
of supt≥0 |x(t)|
. See the following elementary lemma (with φ(t) = 1 + t).
1+t
Lemma 2.1. If φ(t) is continuous on [0, ∞) with inf t≥0 φ(t)= δ > 0 and
lim φ(t)
= 1, then there is K > 0 so that for each x(t) with x ∈ C[0, ∞]
t
t→∞
|x(0)|
|x(t)|
≤
+ K sup |x (t)|
δ
t≥0
t∈[0,∞) φ(t)
sup
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B. Yan and R. M. Timoney
x(t)
t→∞ φ(t)
and lim
= x (∞).
Proof. Let K1 = supt∈[0,∞] |x (t)| and K2 = supt∈[0,∞) t/φ(t). Then
t
|x(0)|
|x(t)|
1 ≤
x
(s)
ds
=
+ K1 K 2 .
x(0)
+
φ(t)
φ(t) δ
0
Given ε > 0, choose T > 0 so that |x (t) − x (∞)| < ε/3 and |1 − t/φ(t)| <
ε/(3K1 ) for all t ≥ T . Then, for t > T
x(t)
− x (∞)
φ(t)
x(T ) − T x (∞)
+
=
φ(t)
t
(x (s) − x (∞)) ds
T
+
φ(t)
t
− 1 x (∞)
φ(t)
and, from the triangle inequality we deduce
x(t)
x(T ) − T x (∞) t − T ε ε
+
+ <ε
φ(t) − x (∞) ≤ φ(t)
φ(t) 3 3
for t > T1 and T1 large enough.
1
From [7] and [8] we know that C[0, ∞] and C∞
[0, ∞) are Banach spaces.
1
We will also consider the cone P of non-negative elements in C∞
[0, ∞)
1
P = {x ∈ C∞
[0, +∞) : x(t) ≥ 0, x (t) ≥ 0 for t ∈ [0, ∞)}.
(2.1)
From the Arzela-Ascoli Theorem (or see [7]) we can state a criterion for
compactness in C[0, ∞].
Lemma 2.2. Let M ⊆ C[0, +∞]. Then M is relatively compact in C[0, +∞]
if and only if it is (norm) bounded in C[0, +∞] and equicontinuous at each
point of [0, ∞]:
(a) sup{x : x ∈ M} < ∞, and
(b) for each t0 ∈ [0, ∞] and ε > 0, there is a neighbourhood U0 of t0 ∈ [0, ∞]
so that t ∈ U0 , x ∈ M implies |x(t) − x(t0 )| < ε.
The next lemma follows from [9, Lemma 2.3.1, Theorem 2.3.2].
Lemma 2.3. Let Ω be a bounded open set in real Banach space E that contains
the origin, P be a convex closed cone in E which is proper (that is P ∩ (−P ) =
{0}) and A : Ω∩P → P a continuous and completely continuous (that is, maps
bounded sets to relatively compact sets) operator. Suppose
λAx = x, ∀x ∈ ∂Ω ∩ P, λ ∈ (0, 1].
(2.2)
Then i(A, Ω ∩ P, P ) = 1 (where i(A, Ω ∩ P, P ) is the fixed point index of A on
Ω ∩ P with respect to P ) and the equation Ax = x has a solution x ∈ Ω ∩ P .
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Positive solutions
Lemma 2.4. Let β ∈ C((0, ∞)) with β(t) < 0 (all t ∈ (0, ∞)), and let μ > 0,
T > 0 and 0 ≤ γ < 1. Suppose x ∈ P satisfies x(T ) > 0, limt→∞ x (t) = 0 and
x (t) ≤ μβ(t)(x(t))γ
(t ∈ [T, ∞)).
Then
x (t) ≥ μ
∞
t
(−β(s))(αT (s))γ ds for t ∈ [T, ∞),
where
1/(1−γ)
t
αT (t) = (1 − γ)μ (τ − T )(−β(τ )) dτ
T
(t ≥ T ≥ 0).
Proof. Note that αT (t) is positive for t > T and satisfies the integral equation
αT (t) = μ
T
t
(τ − T )(αT (τ ))γ (−β(τ )) dτ
Integration from t to +∞ yields that
∞
β(s) (x(s))γ ds,
x (t) ≥ −μ
t
(t ∈ [T, ∞)).
(2.3)
t ∈ [T, +∞).
Thus, for t ≥ T ,
x(t) > x(t) − x(T )
t ∞
≥ −μ
β(τ ) (x(τ ))γ dτ ds
Tt s t
≥ −μ
β(τ ) (x(τ ))γ dτ ds
tT s
≥ μ (τ − T )(−β(τ )) (x(τ ))γ dτ.
T
Note that αT (T ) = 0 < x(T ) and hence there is some interval [T, t∗0 ) with
t∗0 > 1 where x(t) > αT (t) holds. From the inequality
x(t) − αT (t) ≥ μ
T
t
(τ − T ) ((x(τ ))γ − (αT (τ ))γ ) (−β(τ )) dτ
(which follows using (2.3)) we can show that x(t) > αT (t) remains true for all
t ≥ T . This gives the conclusion of the lemma.
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3
B. Yan and R. M. Timoney
Proof of the main result
Lemma 3.1. Let f0 ∈ C([0, +∞) × [0, +∞) × [0, +∞), R), c0 ≥ 0 and c1 ≥ 0.
Assume that for each M > 0 there is ΦM ∈ L1 [0, ∞) so that
x ∈ P, x ≤ M ⇒ |f0 (t, x(t), x (t))| ≤ ΦM (t) (t ≥ 0).
For x ∈ P (P as in (2.1)), we define a function (Ax)(t) (on t ∈ [0, ∞)) by
∞
t
max 0,
f0 (τ, x(τ ), x (τ )) dτ ds.
(Ax)(t) = c0 + c1 x (0) +
0
s
Then Ax ∈ P (for all x ∈ P ) and A : P → P is a continuous and completely
continuous map.
Proof. For x ∈ P , let M = x and then
0 ≤ (Ax)(t)
∞
max 0,
f0 (τ, x(τ ), x (τ )) dτ ds
= (Ax)(0) +
0
s
t ∞
≤ (Ax)(0) +
ΦM (τ ) dτ ds
0
t
0
≤ (Ax)(0) + tΦM 1 < +∞.
(We use · 1 for the integral norm on L1 [0, ∞).) Moreover,
∞
f0 (τ, x(τ ), x (τ )) dτ
0 ≤ (Ax) (t) = max 0,
t
∞
≤
ΦM (τ ) dτ, ∀t ∈ [0, +∞),
t
which yields that limt→∞ (Ax) (t) = 0. Consequently, A : P → P is well
defined.
To show that A : P → P is continuous, we consider a convergent sequence
(xm )∞
m=1 in P converging to x0 ∈ P . By compactness of the sequence and its
limit there is M > 0 so that xm ≤ M for all m (and x0 ≤ M). Because
of the elementary inequality | max{0, u} − max{0, v}| ≤ |u − v| we have
|(Axm ) (t) − (Ax0 ) (t)|
∞
|f0 (τ, xm (τ ), xm (τ )) − f0 (τ, x0 (τ ), x0 (τ ))| dτ
≤
t ∞
≤
|f0 (τ, xm (τ ), xm (τ )) − f0 (τ, x0 (τ ), x0 (τ ))| dτ.
0
(3.1)
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Positive solutions
Also,
|f0 (τ, xm (τ ), xm (τ )) − f0 (τ, x0 (τ ), x0 (τ ))| ≤ 2ΦM (τ )
and (3.1) with the Lebesgue dominated convergence theorem then implies
(Axm ) (t) → (Ax0 ) (t) uniformly for t ≥ 0. One can show in a similar way
that (Axm )(0) → (Ax0 )(0) and then by Lemma 2.1, we have Axm → Ax0 .
Finally, let D ⊆ P be bounded and we claim that A(D) is relatively com1
[0, ∞) since P is closed). Fix M > 0 so that
pact in P (or equivalently in C∞
x ≤ M for x ∈ D. We have
∞
|(Ax) (t)| ≤
ΦM (s) ds
(3.2)
t
∞
ΦM (τ ) ds,
≤
0
which implies that {(Ax) : x ∈ D} is bounded. From (3.2) we see that the
functions (Ax) (x ∈ D) are equicontinuous at ∞. To show equicontinuity at
point t0 ∈ [0, ∞) observe that
∞
f0 (τ, x(τ ), x (τ )) dτ
(Ax) (t) = max 0,
t
t0
= max 0,
f0 (τ, x(τ ), x (τ )) dτ
t
∞
f0 (τ, x(τ ), x (τ )) dτ
+
t0
from which we deduce that
t0
f0 (τ, x(τ ), x (τ )) dτ |(Ax) (t) − (Ax) (t0 )| ≤ t t0
≤ ΦM (τ ) dτ t
and so it follows that {(Ax) (t) : x ∈ D} is equicontinuous at t0 . From
Lemma 2.2 we see that for any (xm )∞
m=1 in D, there is a subsequence (which
∞
we denote again by (xm )m=1 ) so that (Axm ) converges uniformly on [0, ∞] (to
some limit function in z ∈ C[0, ∞]). Passing to a further subsequence we may
assume also that limm→∞ (Axm )(0) = x0 (0) exists. From Lemma
2.1 we can
t
then conclude that limm→∞ Axm = x0 where x0 (t) = x0 (0) + 0 z(τ ) dτ .
Lemma 3.2. Fix f satisfying the hypothesis (1.4) of Theorem 1.1, μ > 0,
a > 0, b ≥ 0, k ≥ 0 and n ∈ N. For x ∈ P (P as in (2.1)), we define a
function (An x)(t) (on t ∈ [0, ∞)) by
∞
t
k b max 0,
−μfn (τ, x(τ ), x (τ )) dτ ds,
(An x)(t) = + x (0) +
a a
0
s
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B. Yan and R. M. Timoney
where we let fn (τ, x, z) = f τ, x + τ +1
, z + n1 . Then An x ∈ P (for all x ∈ P )
n
and An : P → P is a continuous and completely continuous map.
Proof. The idea is to apply Lemma 3.1 with f0 (t, x, z) = −μfn (τ, x, z). What
we need is to establish the existence of ΦM ∈ L1 [0, ∞) for each M > 0 to
satisfy the assumption in Lemma 3.1.
To compress the notation we introduce
Hε,c (τ ) =
max
ε≤x≤c(1+τ )
h(x),
Gε,M = max g(z).
ε≤z≤M
(3.3)
For x ∈ P , it is easy to see that
t
0 ≤ x(t) = x(0) +
x (s)ds ≤ x(0) + tx ≤ (1 + t)x, ∀t ∈ [0, +∞).
0
Thus
1
t+1
≤ x(t) +
≤ (1 + x)(1 + t), ∀t ∈ [0, +∞).
n
n
Fix M > 0 and choose a c > 1 + M big enough such that 1/c < 1/n. Then,
for x ∈ P with x ≤ M,
1
t+1
≤ x(t) +
≤ c(1 + t), ∀t ∈ [0, +∞).
c
n
(3.4)
Thus, using (1.4),
1 τ +1 |fn (τ, x(τ ), x (τ ))| = f τ, x(τ ) +
, x (τ ) +
n
n 1
τ +1
g x (τ ) +
≤ Φ(τ )h x(τ ) +
n
n
τ +1
G1/n,x+1/n .
≤ Φ(τ )h x(τ ) +
n
(3.5)
Together with (3.4) this implies that, for x ∈ P with x ≤ M,
|f0 (τ, x(τ ), x (τ ))| ≤ ΦM (τ ) = μΦ(τ )H1/c,c (τ )G1/n,M +1/n .
As ΦM ∈ L1 [0, ∞) by the hypotheses, we can apply Lemma 3.1 to get the
result.
Lemma 3.3. Fix f satisfying the hypotheses (1.4) and (1.5) of Theorem 1.1,
μ > 0, a > 0, b ≥ 0, k ≥ 0 and n ∈ N. Let An be as in Lemma 3.2. If 1/n < δ,
0 < λ ≤ 1 and x ∈ P satisfies x = λAn x, then
x (t) = λμfn (t, x(t), x (t)) ,
and x (t) > 0 for t ≥ 0.
t ∈ (0, +∞)
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Positive solutions
Proof. Differentiating the formula defining An we have x (t) ≥ 0 (for t ≥ 0),
x(t) ≥ x(0) ≥ 0, and limt→∞ x (t) = 0. Because 1/n < δ, there is t0 > 0 so that
0 ≤ x (t) + 1/n < δ for all t ≥ t0 . From (1.5) we then have fn (t, x(t), x0 (t)) =
f (t, x(t) + (t + 1)/n, x0 (t) + 1/n) ≤ (x(t) + (t + 1)/n)γ β(t) < 0 for t ≥ t0 and
so differentiating the formula defining An gives
∞
fn (τ, x(τ ), x (τ )) dτ
(t ≥ t0 ).
(An x) (t) = −μ
t
Since x = λAn x and fn (τ, x(τ ), x0 (τ )) < 0 (for τ ≥ t ≥ t0 ) we deduce x (t) > 0
for t ≥ t0 .
Thus {t ∈ [0, ∞) : x (s) > 0∀s > t} is nonempty (it contains t0 ) and has
an infimum t∗ . If t∗ > 0 then we must have x (t∗ ) = 0. We will show now that
x (t∗ ) > 0 and this will also show that t∗ = 0, hence x (0) > 0 must hold. If
x (t∗ ) = 0, then by continuity of x we must then have t∗0 > t∗ such that
1
<δ
(t∗ < t ≤ t∗0 ),
n
and hence fn (τ, x(τ ), x0 (τ )) ≤ (x(τ ) + n1 τ + n1 )γ β(τ ) < 0 for t∗ < τ ≤ t∗0 .
As x (t) > 0 for t∗ < t ≤ t∗0 , we have
∞
x (t) = λ(An x) (t) = −μλ
fn (τ, x(τ ), x (τ )) dτ
(3.6)
0 < x (t),
x (t) +
t
∗
t∗0 . In
∗
particular (3.6) holds for t = t∗0 and, by continuity,
in the range t < t ≤
(3.6) also holds at t = t . Thus
t∗0
∗
fn (τ, x(τ ), x (τ )) dτ + x (t∗0 ) > x (t∗0 ) > 0,
0 = x (t ) = −μλ
t∗
a contradiction. We have therefore established t∗ = 0 (and x (0) > 0).
We can now verify directly that x must satisfy
x (t) = λμfn (t, x(t), x (t)) ,
t ∈ (0, +∞).
Notation 3.4. We will continue to use the notation (3.3) with h and g as in
z
u
the hypothesis (1.4). We also let I(z) = Ig (z) = 0 g(u)+1
du. As g(u) > 0
we can see that I is continuous and strictly
on [0, ∞) with I(0) =
∞ increasing
u
0 and I([0, ∞)) = [0, I∞ ) where I∞ = 0 g(u)+1 du ∈ [0, ∞] may be finite
or infinite. Thus there is a monotone increasing continuous inverse function
I −1 : [0, I∞ ) → [0, ∞) with I −1 (0) = 0.
Define, for R > 0.
Φ∞ h(s) + RΦ(s)H1,1+R (s) s ≤ 1
ΨR (s) =
RΦ(s)H1,1+R (s) s > 1.
For μ0 > 0 let J(μ0 ) = {c > 0 : μ0 Ψc 1 < I∞ }.
1198
B. Yan and R. M. Timoney
Lemma 3.5. Assume that f satisfies the hypotheses of Theorem 1.1. Then,
using the notation above, there is μ0 > 0 so that
sup
k
c∈J (μ0 ) a
c
> 1.
+ ( ab + 1)I −1 (μ0 Ψc 1 )
(3.7)
Moreover, for such μ0 there is R1 > 0 and ε > 0 so that
k
b
R1 > +
+ 1 I −1 (I(ε) + μ0 ΨR1 +ε 1 ) .
a
a
(3.8)
Proof. Fix R > k/a and then we can find μ0 > 0 so that
μ0 Φ∞
1
h(s) ds + R
0
∞
0
Φ(s)H1,1+R (s) ds
< I∞
and
k
a
R
> 1.
∞
1
b
−1
μ0 Φ∞ 0 h(s) ds + R 0 Φ(s)H1,1+R (s) ds
+ ( a + 1)I
We have μ0 so that (3.7) holds.
Then there is R > 0 so that R > (k/a) + (b/a + 1) I −1 (μ0 ΨR 1 ) .
Further, by continuity of I and its inverse, we can choose ε > 0 so that
b
k
R −ε > +
+ 1 I −1 (I(ε) + μ0 ΨR 1 )
a
a
and ε < R . Then let R1 = R − ε.
Lemma 3.6. Assume that f satisfies the hypotheses of Theorem 1.1 and fn
is as in Lemma 3.2. Assume also that 0 < μ ≤ μ0 , R1 > 0 where μ0 , R1 and
ε > 0 satisfy (3.8) and n ≥ n0 > 1/ min{ε, δ}. Then the problem
x (t) = μfn (t, x(t), x (t)) ,
k b + x (0) ,
x(0) =
a a
t ∈ (0, +∞)
x (∞) = 0
(3.9)
(3.10)
has a solution x = xn ∈ P with 0 < xn < R1 and xn (t) > 0 for all t ≥ 0.
Proof. We will consider the continuous and completely continuous operator
An of Lemma 3.2 (for n ≥ n0 ) and we aim to apply Lemma 2.3 to each An
1
where Ω = {x ∈ C∞
[0, +∞) : x < R1 }. To that end, we claim that there is
no x ∈ P ∩ ∂Ω = {x ∈ P : x = R1 } with x = λAn x and 0 < λ ≤ 1.
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Positive solutions
Assume on the contrary that the claim is false and x = x0 ∈ P , 0 < λ0 ≤
1, n ≥ n0 satisfy x0 = R1 and x0 = λ0 An x0 . By Lemma 3.3, x (t) =
λ0 μfn (t, x(t), x (t)) (for t > 0). Thus, by (1.4),
t+1 1 , x0 (t) +
−x0 (t) ≤ μ f t, x0 (t) +
n
n t+1
g(x0 (t) + 1/n);
≤ μΦ(t)h x0 (t) +
n
− (x0 (t) + 1/n) x0 (t)
t+1
≤ μΦ(t)h x0 (t) +
(x0 (t) + 1/n).
g(x0 (t) + 1/n) + 1
n
Integrating between t and ∞ we get
I(x0 (t) + 1/n) − I(1/n)
∞
s+1
s+1
d x0 (s) +
≤ μ
Φ(s)h x0 (s) +
n
n
t
∞
s+1
s+1
≤ μ
Φ(s)h x0 (s) +
d x0 (s) +
.
n
n
0
(3.11)
Since limt→+∞ (x0 (t) + (t + 1)/n) = +∞ and x0 (t) + (t + 1)/n is increasing,
≥ 1 for all t ∈ [0, +∞); or x0 (0)+ n1 <
there are two situations: either x0 (t)+ t+1
n
1.
In the second case there is a unique T > 0 such that x0 (T ) + n1 T + n1 = 1
and 1 ≤ x0 (t) + n1 t + n1 ≤ (R1 + ε)(1 + t), ∀t ∈ [T, +∞). And integrating
between t and ∞ we estimate (3.11) from above by
I(x0 (t) + 1/n) − I(1/n)
T
s+1
s+1
Φ(s)h x0 (s) +
≤ μ
d x0 (s) +
n
n
0
∞
s+1
s+1
Φ(s)h x0 (s) +
+μ
d x0 (s) +
n
n
T
T s+1
s+1
h x0 (s) +
d x0 (s) +
≤ μΦ∞
n
n
0
∞
s+1
1
Φ(s)h x0 (s) +
x0 (s) +
ds
+μ
n
n
T
∞
1
h(s) ds + (R1 + ε)
Φ(s)H1,R1 +ε (s) ds
≤ μ Φ∞
= μΨR1 +ε 1
0
0
(∀t ∈ [0, +∞)). In the first case we get the same estimate by taking T = 0 in
the above argument.
1200
B. Yan and R. M. Timoney
Rearranging, and using I(1/n) < I(ε) together with monotonicity of I −1
we get
I(x0 (t) + 1/n) ≤ I(ε) + μΨR1 +ε 1
x0 (t) ≤ I −1 (I(ε) + μΨR1 +ε 1 ) < R1
(3.12)
by (3.8). Integrating again from 0 to t, we find
t
x0 (t) = x0 (0) +
x0 (s) ds ≤ x0 (0) + tI −1 (I(ε) + μΨR1 +ε 1 )
0
from which it follows that
sup
t≥0
|x0 (t)|
≤ x0 (0) + I −1 (I(ε) + μΨR1 +ε 1 )
1+t
k
b
≤ λ0 + λ0 x0 (0) + I −1 (I(ε) + μΨR1 +ε 1 )
a a b
k
≤
+
+ 1 I −1 (I(ε) + μΨR1 +ε 1 ) < R1
a
a
(3.13)
by (3.8). From (3.12), (3.13) and 1/n < ε we get R1 = x0 < R1 , a contradiction. This establishes the claimed nonexistence of x0 .
We can now apply Lemma 2.3 to An to obtain xn ∈ P ∩ Ω with xn = An xn .
By Lemma 3.3, x = xn must satisfy (3.9)–(3.10) and xn (t) > 0 for t ≥ 0.
Proof (of Theorem 1.1). Let n0 be as in Lemma 3.6. For n ≥ n0 , let xn ∈ P
satisfy (3.9)–(3.10) and xn (t) > 0 for all t ≥ 0. Let Mn = supt∈[1,+∞) xn (t) and
η = inf n≥n0 Mn . Now we show that η > 0. In fact, if η = 0, there is an {nj }
such that Mnj + n1j → 0 as j → +∞ and we can assume Mnj + n1j < δ (all j).
Our assumption (1.5) implies that
γ
t+1
fnj (t, xnj (t), xnj (t)) ≤ β(t) xnj (t) +
, t ∈ [1, +∞).
nj
Thus,
xnj (t)
γ
t+1
≤ μβ(t) xnj (t) +
, t ∈ [1, +∞)
nj
and hence
xnj (t) ≤ μβ(t)(xnj (t))γ
By Lemma 2.4
xnj (t)
≥μ
t
∞
t ∈ [1, +∞).
(−β(s))(α1 (s))γ ds, t ∈ [1, +∞)
1201
Positive solutions
and this contradicts η = 0.
Let ζ = min{δ/2, η}. Since limt→+∞ xn (t) = 0, set xn (tn ) = ζ, tn ∈
[1, +∞). It is easy to see that xn (t) is decreasing on [tn , +∞). Now we show
that
xn (t) ≥ ζ, t ∈ [0, tn ], n ≥ n0 .
(3.14)
In fact, if there is a t0 ∈ [0, tn ] with xn (t0 ) < ζ, set t∗ = sup{t|xn (s) < ζ for all
s ∈ [t0 , t]}. Obviously, t∗ ≤ tn with xn (t) < ζ for all t ∈ [t0 , t∗ ) and xn (t∗ ) = ζ.
Now (1.5) implies that
γ
t+1
xn (t) ≤ μβ(t) xn (t) +
, t ∈ [t0 , t∗ ],
n
which yields that xn (t) is decreasing on [t0 , t∗ ]. This is a contradiction to
xn (t∗ ) = ζ > xn (t0 ) and establishes (3.14).
t
Since xn (t) ≥ ζ for all t ∈ [0, tn ] ⊇ [0, 1], xn (t) = xn (0) + 0 xn (s) ds ≥ ζt
for all t ∈ [0, 1]. Since xn (t) is increasing on [0, +∞), we know that xn (t) ≥ ζ
for all t ∈ [1, +∞), n ∈ N0 . Then,
t+1
≤ (R1 + ε)(1 + t), t ∈ [1, +∞).
n
From (3.9) and (1.4) we get
ζ ≤ xn (t) +
−(xn (t) + 1/n)xn (t)
(xn (t) + 1/n)|xn (t)|
≤
g(xn (t) + 1/n) + 1
g(xn (t) + 1/n) + 1
≤ μΦ(t)(xn (t) + 1/n)h(xn (t) + (t + 1)/n)) (t > 0)
and integrating the inequality gives
|I(xn (t1 ) + 1/n) − I(xn (t2 ) + 1/n)|
t2
t
+
1
1
Φ(t)h xn (t) +
≤ μ xn (t) +
dt .
n
n
t1
(3.15)
(for 0 ≤ t1 , t2 < ∞). Then we have for t1 ≥ 1 and t2 ≥ 1
t2
(R1 + ε)Φ(t)Hζ,R1 +ε (t) dt .
|I(xn (t1 ) + 1/n) − I(xn (t2 ) + 1/n)| ≤ μ t1
(3.16)
If t1 ≤ 1 and t2 ≤ 1
|I(xn (t1 ) + 1/n) − I(xn (t2 ) + 1/n)|
t2
t+1
t+1
Φ(t)h xn (t) +
≤ μ
d xn (t) +
dt
n
n
t1
xn (t2 )+(t2 +1)/n
h(s) ds
≤ μ Φ∞
xn (t1 )+(t1 +1)/n
(3.17)
1202
B. Yan and R. M. Timoney
∞
(3.16) and finiteness of 1 Φ(s)H1/c,c (s) ds guarantee that the functions
I(xn (t) + 1/n) (n ≥ n0 ) are equicontinuous at each point of [1, ∞). Since
{xn (t) + 1/n : n ≥ n0 } is bounded on [0, 1] by supn xn + 1/n0 ≤ R1 + 1, the
functions belonging to {xn (t) + (t + 1)/n : n ≥ n0 } are equicontinuous on [0, 1].
T
Together with (3.17) and the finiteness of 0 h(s) ds (for T = 2R1 +2 > 0) this
implies that the functions I(xn (t) + 1/n) (n ≥ n0 ) are equicontinuous at each
point of [0, 1]. Combining with equicontinuity on [1, ∞) we get equicontinuity
at each point of [0, ∞). Since I −1 is uniformly continuous on [0, I(R1 + 1/n0 )],
we can deduce that the functions xn (t) (n ≥ n0 ) are equicontinuous at each
point of [0, ∞). A similar argument, using xn (∞) = 0 shows, via
+∞
Φ(t)Hζ,R1 +ε (t) dt (t1 ≥ 1),
|I(xn (t1 ) + 1/n) − I(1/n)| ≤ μ t1
that the functions are also equicontinuous at ∞.
We can therefore apply Lemma 2.2 to conclude that there is a subsequence
∞
(xnj )∞
j=1 of (xn )n=n0 that converges uniformly on [0, ∞] to some limit function
y ∈ C[0, ∞]. Passing to a further subsequence, and using compactness of
[0, R1 ] we can also assume that limj→∞ xnj (0) = ξ0 exists. We can then define
1
x0 ∈ C∞
[0, +∞) by
t
x0 (t) = ξ0 +
y(τ ) dτ
0
and we can see from the fact that xn satisfies (3.10) that x = x0 must satisfy
the boundary conditions (1.2). As xnj ∈ P we have x0 (0) = ξ0 ≥ 0. As
xnj (t) > 0 for all t we have x0 (t) = y(t) = limj→∞ xnj (t) ≥ 0.
By continuity at ∞ and x0 (∞) = 0, there is t0 > 0 so that x0 (t) < δ/3 for
t ∈ [t0 , ∞). By uniform convergence, there is j0 > 3/δ so that |xnj (t)−x0 (t)| <
δ/3 for all j ≥ j0 and t∈ [t0 , ∞). Hence 0 < xnj (t) + 1/nj < δ for all t ≥ t0 ,
j ≥ j0 . From (1.5) we have therefore
γ
τ +1
β(τ ) (τ ≥ t0 , j ≥ j0 ).
fnj (τ, xnj (τ ), xnj (τ )) ≤ xnj (τ ) +
nj
Since xn satisfies (3.9), Lemma 2.4 gives
∞
(αt0 (τ ))γ (−β(τ )) dτ
xnj (t) ≥ μ
t
(t ≥ t0 , j ≥ j0 ).
Taking limits as j → ∞ we see that x0 (t) > 0 holds for all t ∈ (t0 , ∞). We
claim that in fact x0 (t) > 0 for all t ≥ 0. To see the claim, consider the set
{t ≥ 0 : x0 (τ ) > 0∀τ ∈ (t, ∞)} ⊆ [0, ∞), a nonempty set since it contains t0 .
Denote the infimum of this set by t∗ . If t∗ > 0 then necessarily x0 (t∗ ) = 0.
1203
Positive solutions
We show that x0 (t∗ ) > 0 (so that t∗ = 0 and x0 (0) > 0). If x0 (t∗ ) = 0, then
there is t1 > t∗ so that x0 (t) < δ/3 for t∗ ≤ t ≤ t1 . By unform convergence
there is j0 ≥ 3/δ such that |xnj (t) − x0 (t)| < δ/3 for j ≥ j0 and all t. Hence
xnj (t) + 1/nj < δ for t ∈ [t∗ , t1 ) and j ≥ j0 . For j ≥ j0 fixed, consider
t∗j = sup{tj : xnj (t) + 1/nj < δ∀t ∈ [t∗ , tj )} ≥ t1 > t∗ . Note that
xnj (t) = μfnj (t, xnj (t), xnj (t))
γ
t+1
β(t) < 0 (t∗ ≤ t < t∗j )
< μ xnj (t) +
nj
(3.18)
by (1.5). If t∗j < ∞, then xnj (t∗j ) = δ + 1/nj , but this is impossible because
(3.18) implies δ > xnj (t∗ ) + 1/nj > xnj (t∗j ) + 1/nj = δ. Thus t∗j = ∞ for j ≥ j0 .
From Lemma 2.4 we have
∞
xnj (t) ≥ μ
(αt∗ (τ ))γ (−β(τ )) dτ (t ≥ t∗ , j ≥ j0 ).
t∗
Taking the limit as j → ∞ shows x0 (t∗ ) > 0. We have therefore shown that
t∗ = 0 and x0 (t) > 0 for t ≥ 0, which implies that x0 (t) > 0 for all t ∈ (0, +∞).
Consequently, since limj→+∞ xnj − x0 = 0, we have
inf
j≥1
and
min xnj (s), min
xnj (s)
min xnj (s), min
xnj (s)
s∈[t,1]
s∈[t,1]
inf
j≥1
> 0, t < 1
s∈[1,t]
s∈[1,t]
> 0, t > 1.
From
xnj (t)
−
xnj (1)
t
=μ
1
fnj (τ, xnj (τ ), xnj (τ )) dτ, t ∈ (0, +∞),
letting j → +∞, the Lebesgue Dominated Convergence Theorem and (1.4)
guarantee that
t
x0 (t) − x0 (1) = μ
f (τ, x0 (τ ), x0 (τ )) dτ, t ∈ (0, +∞).
1
By direct differentiation, we have
x0 (t) = μf (t, x0 (t), x0 (t)), t ∈ (0, +∞),
which is (1.1). We already established that x = x0 satisfies (1.2).
1204
4
B. Yan and R. M. Timoney
Examples and consequences
Corollary 4.1. Assume that the hypotheses of Theorem 1.1 hold together with
one of the following conditions
∞
(i) limt→∞ t t β(τ ) dτ = −∞
∞
(ii) t0 (τ − t0 )β(τ ) dτ = −∞ for some t0 ≥ 0.
Then all solutions of (1.1)–(1.2) with x(t) ≥ 0 and x (t) > 0 for t ∈ (0, ∞)
are unbounded.
Proof. Let x0 be a nonnegative and strictly monotone increasing solution of
(1.1)–(1.2). Since limt→∞ x0 (t) = 0, there is t ≥ t0 so that 0 < x (t) < δ for
all t ∈ [t , ∞). From (1.5) and the integral form of (1.1) we have
x0 (t)
≥μ
∞
t
(x0 (τ ))γ (−β(τ )) dτ
(t ≥ t ).
Hence, for t > t ,
t
x0 (t) − x0 (t ) ≥ μ
t
t s t
= μ
t
= μ
∞
(x0 (τ ))γ (−β(τ )) dτ ds
(x0 (τ ))γ (−β(τ )) dτ ds
s
t ∞
(x0 (τ ))γ (−β(τ )) dτ ds
+μ
t
t
t
(τ − t )(x0 (τ ))γ (−β(τ )) dτ
t
∞
(x0 (τ ))γ (−β(τ )) dτ.
+μ(t − t )
t
Given either of the two conditions we conclude x(t) → ∞ as t → ∞.
Corollary 4.2. Assume that the hypotheses of Theorem 1.1 hold and that for
each c > 0, c ≥ 1 there is T = T (c , c) ∈ [1, ∞) satisfying
T
∞
I −1 c
∞
s
Φ(τ ) 1 max
c
≤x≤c(1+τ )
h(x) dτ
ds < ∞
z
(with I(z) = 0 u/(g(u) + 1) du). Then all solutions of (1.1)–(1.2) with x(t) ≥
0 and x (t) > 0 for t ∈ (0, ∞) are bounded.
1205
Positive solutions
Proof. Let x0 be a nonnegative and strictly monotone increasing solution of
(1.1)–(1.2). From the assumptions limt→0+ x0 (t) exists and limt→∞ x0 (t) = 0.
1
It follows easily that x0 ∈ C∞
[0, ∞). As in (3.4) there is a c > 0 such that
1
≤
x
(t)
≤
c(1
+
t)
for
all
t
∈
[1,
+∞). Then, we can see from (1.1) and (1.4)
0
c
that
|x0 (t)| ≤ μΦ(t)h(x0 (t))g(x0 (t))
−x0 (t)x0 (t)/(g(x0 (t)) + 1) ≤ μΦ(t) 1 max h(x)x0 , t ∈ [1, +∞).
c
≤x≤c(1+t)
Integrating we get
I(x0 (t))
− I(0) ≤ μx0 t
∞
Φ(τ ) 1 max
≤x≤c(1+τ )
c
h(x) dτ, t ∈ [1, +∞),
and hence (since I(0) = 0 and I −1 is monotone) if t ∈ [t0 , ∞) for t0 ≥ 1
sufficiently large
∞
x0 (t) ≤ I −1 μx0 Φ(τ ) 1 max h(x) dτ .
t
c
≤x≤c(1+τ )
Integrating both sides and using the hypothesis we get x0 bounded.
Example 4.3. Consider the boundary value problem
x = μe−t xb1 + xb2 + x−b3 (cos t + (x )a1 − (x )−a2 ) (t ∈ (0, ∞))
limt→∞ x (t) = 0
x(0) − x (0) = 0,
(4.1)
where a1 ≥ 0, a2 > 0, b2 ≥ b1 > 0, 0 < b3 < 1, b1 < 1 and μ > 0. Then there
is μ0 > 0 so that, for 0 < μ < μ0 , the problem has a bounded nonnegative
strictly monotone increasing solution. If b2 + 1 < 2 − a1 , μ0 = +∞.
To see that this is so we apply Theorem 1.1 and Corollary 4.2, where
k = 0, a = b = 1, Φ(t) = e−t , h(x) = xb1 + xb2 + x−b3 , g(z) = 1 + z a1 + z −a2 ,
δ = (1/3)1/a2 , γ = b1 and β(t) = −e−t . It is easy to see that (1.4) and (1.5)
hold. Theorem 1.1 guarantees that there is μ0 > 0 so that for 0 < μ < μ0 the
problem has a nonnegative strictly monotone increasing solution.
1
Moreover, since limu→0+ 1+ua2 +a
a1 +a2 = 1, there is a δ0 with 1 ≥ δ0 > 0
u1+a2
such that for 0 ≤ u ≤ δ0 , 1+ua2 +ua1 +a2 ≥ 12 u1+a2 . Thus
z
I(z) =
0 z
≥
0
z
u
u1+a2
du =
du
a2
a1 +a2
1 + g(u)
0 1 + 2u + u
1 1+a2
1
u
z 2+a2 (0 ≤ z ≤ δ0 ),
du =
2
2(2 + a2 )
1206
B. Yan and R. M. Timoney
which implies
1
I −1 (y) ≤ (2(2 + a2 )y) 2+a2 ,
0 ≤ y ≤ I(δ0 ).
(4.2)
∞
Since lims→+∞ s Φ(τ )H1/c,c (τ ) dτ = 0 (with the notation (3.3)), there is a
∞
T = T (c , c) > 0 such that max(c , 1) T Φ(τ )H1/c,c (τ ) dτ < I(δ0 ). Then (4.2)
yields that
∞
∞
−1
I
Φ(τ )H1/c,c (τ ) dτ ds
c
T
≤
T
s
∞
2(2 + a2 )c
s
1
2+a
∞
Φ(τ )H1/c,c (τ ) dτ
2
ds < +∞.
As a result Corollary 4.2 guarantees that the solutions to (4.1) are bounded.
Assume now b2 + 1 < 2 − a1 . For z ≥ 1,
z
u
I(z) =
du
0 1 + g(u)
z
1
u
u
du +
du
=
0 1 + g(u)
1 1 + g(u)
1
≥ I(1) +
[z 2−a1 − 1],
3(2 − a1 )
we have I([0, ∞)) = [0, ∞) and
−1
I(1) +
z≥I
1
2−a1
[z
− 1] , z ≥ 1.
3(2 − a1 )
1
[z 2−a1 − 1], so that y ≥ I(1) can be arbitrary with z ≥ 1.
Let y = I(1) + 3(2−a
1)
Hence
1
I −1 (y) ≤ (3(2 − a1 ) (y − I(1)) + 1) 2−a1
(y ≥ I(1)).
(4.3)
Finally, for any μ0 > 0, consider (3.7). We have already noted I∞ =
∞ and so J(μ0 ) = [0, ∞). It is easy to see that, with the constant c0 =
1
Φ∞ 0 h(s) ds,
∞
k
b
−1
μ 0 c0 + c
+
+1 I
Φ(s)
max
h(x) ds
1≤x≤(1+c)(1+s)
a
a
0
∞
−1
b2
−s
b2
e (1 + s) ds
.
μ0 c0 + 3c(1 + c)
≤ 2I
0
From (4.3), it is easy to see that if b2 + 1 < 2 − a1 , then the left hand side of
(3.7) is ∞ for all μ0 . Thus (4.1) has solutions for all μ (if b2 + 1 < 2 − a1 ).
Positive solutions
Example 4.4. Consider the boundary value problem (on t ∈ (0, ∞))
x = μ(1 + t)−2 xb1 + xb2 + x−b3 (1 + (x )a1 − (x )−a2 )
x(0) = 0,
limt→∞ x (t) = 0,
1207
(4.4)
where 1 ≥ a1 ≥ 0, a2 > 0, 1 > b1 ≥ 0, 1 > b2 > 0, 1 > b3 ≥ 0 and μ > 0. Then
there is a μ0 > 0 such that the problem (4.4) has an unbounded nonnegative
strictly monotone increasing solution for all 0 < μ ≤ μ0 .
To see that this is so we apply Theorem 1.1, Corollary 4.1, where k = 0,
a = 1, b = 0, Φ(t) = (1 + t)−2 , h(x) = xb1 + xb2 + x−b3 , g(z) = 1 + z a1 + z −a2 ,
δ = (1/3)1/a2 , γ = b1 and β(t) = −(1 + t)−2 .
∞
To apply Corollary 4.1 note that 1 (τ − 1)β(τ ) dτ = −∞.
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Received: February 14, 2007